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Math Help - Fibonacci Look-a-like

  1. #1
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    Fibonacci Look-a-like

    Define a fibonacci series generated by {a,b} such as,
    u_1=a
    u_2=b
    u_{n+2}=u_{n+1}+u_{n}
    (The original fibonacci series is generated by {1,1})
    ( a\not =0)

    Then there are a number of interesting properties:
    1)If K is the finite continued fraction for b/a
    then \frac{u_{n+1}}{u_n}=[1;1,1...,1,K]
    Where the 1 appears n-2 times.

    2)Thus, from here we have that \lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\psi
    (Thus, any fibonacci series generated by any two real numbers converges to the divine proportion).

    3)The formula for u_n is given by
    u_n=F(n-2)a+F(n-1)b
    where F(n) is the n-th fibonacci number.
    But by Binet's formula we have that,
    for the n-th fibonacci number we can find a formula for u_n but it is rather messy and will be omitted. Giving a second method for proving statement 2.
    Last edited by ThePerfectHacker; November 22nd 2006 at 10:53 AM.
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  2. #2
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    I think you mean \phi and not \psi. Very interesting, though.
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  3. #3
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    Quote Originally Posted by Treadstone 71
    I think you mean \phi and not \psi. Very interesting, though.

    <br />
\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi<br />
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  4. #4
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    Quote Originally Posted by Natasha1
    <br />
\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi<br />
    That is not important
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  5. #5
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    The proof of the second statement is rather neat.

    2)\;\;\lim_{n\to\infty}\frac{u_{n+1}}{u_n}\:=\: \phi

    The limit of the ratio of consecutive Fibonacci numbers is already established.

    . . . . . [1]\;\lim_{n\to\infty}\frac{F_{n+1}}{F_n}\:=\:\phi \qquad\qquad[2]\;\lim_{n\to\infty}\frac{F_n}{F_{n+1}} \:=\:\frac{1}{\phi}


    We have: u_n\:=\:a\!\cdot\!F(n-2) + b\!\cdot\!F(n-1)

    The ratio is: R\;=\;\frac{u_{n+1}}{u_n}\,=\,\frac{a\!\cdot\!F(n-1) + b\!\cdot\!F(n)}{a\!\cdot\!F(n-2) + b\!\cdot\!F(n-1)}

    Divide top and bottom by F_{n-1}:\;\;R\;=\;\frac{a + b\!\cdot\!\frac{F(n)}{F_{n-1)}}}{a\!\cdot\!\frac{F(n-2)}{F(n-1)} + b}

    Take the limit: \lim_{n\to\infty}R\;=\;\lim_{n\to\infty}\left( \frac{a + b\!\cdot\!\frac{F(n)}{F_{n-1)}}}{a\!\cdot\!\frac{F(n-2)}{F(n-1)} + b}\right)

    From [1] amd [2], this becomes: \frac{a + b\!\cdot\!\phi}{a\!\cdot\!\frac{1}{\phi} + b}

    Multiply top and bottom by \phi:\;\;\frac{\phi(a + b\!\cdot\!\phi)}{a + b\!\cdot\!\phi}\;=\;\phi

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  6. #6
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    Quote Originally Posted by Soroban
    The proof of the second statement is rather neat.

    [size=3]
    My proof is simpler.
    Because of condition 1), since this countinued fractions converges to [1:1,1,...]=\phi.
    Last edited by ThePerfectHacker; June 4th 2006 at 06:40 AM.
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  7. #7
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    Hello, ThePerfectHacker!

    My proof is simpler.
    Of course it is . . . I wasn't claiming otherwise.

    I was demonstrating a straight algebraic approach to the limit
    . . for those not familiar with continued fractions.

    Please understand: I never ever compete with other posters here.
    I may post an alternate approach or even an improvement
    . . . but never with a snicker or a sneer, implied or otherwise.
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